GATE Overflow Test Series | Mock GATE | Test 2 | Question: 65

Question

Which of the following is/are correct formalization of the sentence? (Mark all the appropriate choices)

" Every student will do well in some examination"

Here, $dowell(x,y)$ means student $x$ will do well in examination $y.$

• A. $\forall x(student(x) \implies \exists y(exam(y) \wedge dowell(x,y)))$
• B. $\forall x(student(x) \implies \exists y(exam(y) \implies dowell(x,y)))$
• C. $\neg \exists x(student(x) \implies \exists y(exam(y) \wedge dowell(x,y)))$
• D. $\neg \exists x(student(x) \wedge \neg \exists y(exam(y) \implies dowell(x,y)))$

Answer 1

Option A is the  only correct representation of the given sentence.

• A. $\forall x(student(x) \implies \exists y(exam(y) \wedge dowell(x,y)))$
  For every student there is an examination in which he will do well.
• B. $\forall x(student(x) \implies \exists y(exam(y) \implies dowell(x,y)))$
  For every student there is a non examination (becomes trivially true) or an examination in which where he will do well.
• C. $\neg \exists x(student(x) \implies \exists y(exam(y) \wedge dowell(x,y)))$
  Everyone is a student (obviously false) or there exist a non-student who will do well in some examination
• D. $\neg \exists x(student(x) \wedge \neg \exists y(exam(y) \implies dowell(x,y)))$
  There does not exist a student such that, either everything is an examination (obviously false) or there exists a non-exam in which he will do well

In examination one can easily identify that options B,C and D are not correct even without fully getting the English meaning.

Comments

That’s right @viniit. But Do Not assume some domain unless it is mentioned. By default, assume that the domain contains everything.

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For Option D

$\neg \exists{x}(student(x) \land \neg \exists y(exam(y) \implies dowell(x,y))) $

$\forall{x}(\neg student(x) \lor \neg \neg \exists y(exam(y) \implies dowell(x,y)))$

$\forall{x}(\neg student(x) \lor \exists y(exam(y) \implies dowell(x,y))) $

$ \forall{x}(student(x) \implies \exists y(exam(y) \implies dowell(x,y)))$

@Deepak Poonia, sir what is wrong with this?

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@faisal_sayyed IIUC, I believe there is a difference between the two statements ‘there exists \(y\) such that if \(y\) is an exam, then \(x\) will do well in exam \(y\)’ and ‘there exists \(y\) such that \(y\) is an exam, and \(x\) will do well in exam \(y\)’ (which is what the question originally demands of, the existence of such an exam). Generally, we use \(\wedge\) when using \(\exists\) as a quantifier, and \(\Rightarrow\) when using \(\forall\) as a quantifier in a FoL statement.